The coordinate version of the exterior derivative $d:\Omega^k(M)\to \Omega^{k+1}(M)$ of differential forms of a $C^\infty$ manifold $M$ can be expressed on a form $$ \omega=fdx^1\wedge\cdots\wedge dx^k$$ as $$ d\omega=\sum_{i=1}^n\frac{\partial f}{\partial x^i}dx^i\wedge dx^1\wedge\cdots \wedge dx^k$$ (for example). Alternatively, one can express this in a coordinate invariant form as $$ d\omega(X_1,\ldots, X_{k+1})=\sum_{i=1}^{k+1}(-1)^{i+1}X_i(\omega(X_1,\ldots, \widehat{X_i},\ldots, X_{k+1}))$$ $$+\sum_{1\le i<j\le k+1}(-1)^{i+j}\omega([X_i,X_j],X_1,\ldots, \widehat{X_i},\ldots, \widehat{X_j},\ldots,X_{k+1}).$$ Here the $X_i\in \mathfrak{X}(M)$. I can show that these two operators are actually the same. That's not a problem. My confusion is in showing that the coordinate invariant form $d\omega$ is independent of extension $X_1,\ldots, X_{k+1}\in \mathfrak{X}(M)$. Indeed, the idea is that we define $d\omega_p$ on a $(k+1)-$tuple $$(v_1,\ldots, v_{k+1})\in \overbrace{T_pM\times\cdots\times T_pM}^{k+1\:\text{times}}$$ by extending this tuple to a $(k+1)-$tuple of smooth vector fields $X_1,\ldots, X_{k+1}$ so that $X_i(p)=v_i$. I can see that this is independent of extension, because of the corresponding fact for the coordinate definition.
On the other hand, I would like to show that this formula is well-defined in this sense intrinisically, based only on the coordinate independent formula.
EDIT: As an addendum, I want to understand why the coordinate independent definition is independent of extension $(X_1,\ldots, X_{k+1})$ chosen without appealing to the coordinate definition.
EDIT 2: Please note this question is not a duplicate of the other, because I am interested in showing the corresponding fact for the coordinate-independent version of the formula, without appealing to the coordinate expression.